Page 416 - Analog and Digital Filter Design

Design Equations 41

It follows from this that at w= 1, H(w) = 1/4, or 0.7071. This is -3dB relative
to the zero frequency point, no matter what value of filter order is considered,
sinse 1" = 1. For this reason the -3 dB cutoff frequency is considered the natural
passband edge. Attenuation for any filter order at other frequencies can be found
by substituting different values of o and PZ.
For example, at w = 3 and IZ = 5. Attenuation = lO.log(l/(l + 3")) = 47.7dB,
which agrees with the graph in Figure 2.10. Another way of looking at this is
to find the filter order that will satisfy the attenuation requirements.

In A
I2 = -
In R
-1
-1
R=w,/w,,

In the case of the normalized 3dB filter, K,, = 3dB so loo''/' = 2 and A =
~~~
d(10"'"' - 1). The passband frequency is w/, = 1 and so R = w,.
For example, to find the filter order required for a normalized filter with 65dB
attenuation at w = 2.5:

A=1778 and R=3.5
1n1778
!I=-- -5.1669
ln2.5
An eighth-order filter is just outside the attenuation limit, so a ninth-order
design is necessary.

Butterworth Phase

To find the phase response of Butterworth filters it is necessary to expand the
transfer function, to give a denominator with a polynomial.

1
H(s) =
a(, + a, s + as + cI;s-' + . . .
Fortunately there is an iterative equation that can be used to find the poly-
nomial coefficients, a,<. The initial coefficient, a. = 1.

sos[(k - 1)7c/2PZ]
a/> = k=l,2,3, ... ?I
sin(kn: 2n)